Question

Определить длину волны лучей ,кванты которых имеют такую же энергию,что и электрон, пролетевшии разность потенциалов 4,1В . С решение и ДАНО

Answer 1

Помоему, для решения задачи данных не хваатет, а имменно работы выхода. 

Воспользуемся уравнением энштейна для фотоэффекта:

Е(фотона)=Е(выхода) + еU(задерживающ)

hc/лямбда = Е(выхода) + еU(задерж)

лямбда = Hc/(Е(выхода) + еU(задерж))

Answer 2

Calculating the wavelength of photons with the same energy as an electron accelerated through a 4.1 V potential difference

To solve this problem, we will use the formula that relates the kinetic energy of a particle to the potential difference it was accelerated through:

`KE = qV`

Where: - KE is the kinetic energy of the particle - q is the charge of the particle (in coulombs) - V is the potential difference (in volts)

We are given that the potential difference is 4.1 V. The charge of an electron is -1.6×10^-19 C.

Plugging these values into the formula gives:

`KE = (-1.6×10^-19 C) × 4.1 V = 6.56×10^-19 J`

This is the kinetic energy that the electron gained by passing through the 4.1 V potential difference.

We are also given that this kinetic energy should be equal to the energy of photons. The energy of a photon is given by:

`E = hf`

Where: - E is the energy of the photon - h is Planck's constant (6.626×10^-34 J·s) - f is the frequency of the photon

Setting the kinetic energy of the electron equal to the energy of the photon and solving for the frequency f gives:

`6.56×10^-19 J = (6.626×10^-34 J·s)f`

`f = 9.89×1014 Hz`

Finally, the wavelength λ of a photon is related to its frequency by:

`c = λf`

Where c is the speed of light (3×108 m/s).

Plugging in the values gives:

`λ = c/f = 3×108 m/s / 9.89×1014 Hz = 3.03×10^-7 m = 303 nm`

Therefore, the wavelength of photons with the same energy as the electron accelerated through the 4.1 V potential is 303 nm.